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Let's suppose I have a finite potential well: $$
V(x)=
\begin{cases}
\infty,\quad x<0\\
0,\quad 0<x<a\\
V_o,\quad x>a.
\end{cases}
$$
I solved the time-independent Schrodinger equation for each region and after applying the continuity conditions of ##\Psi## and its derivative I ended up with:
$$ \tan(k_1a)=-\frac{k_1}{k_2},$$ where ##k_1=\sqrt{\frac{2mE}{\hbar^2}}## and ##k_2=\sqrt{\frac{2m(V_o-E)}{\hbar^2}}##.
I'm aware of the fact that solutions can only be calculated graphically, but what's the relation between the value of ##V_o## and the bound states? What if I want to find the acceptable values of ##V_o## for the bound states to be ##1,2,3,\dots## or none?
V(x)=
\begin{cases}
\infty,\quad x<0\\
0,\quad 0<x<a\\
V_o,\quad x>a.
\end{cases}
$$
I solved the time-independent Schrodinger equation for each region and after applying the continuity conditions of ##\Psi## and its derivative I ended up with:
$$ \tan(k_1a)=-\frac{k_1}{k_2},$$ where ##k_1=\sqrt{\frac{2mE}{\hbar^2}}## and ##k_2=\sqrt{\frac{2m(V_o-E)}{\hbar^2}}##.
I'm aware of the fact that solutions can only be calculated graphically, but what's the relation between the value of ##V_o## and the bound states? What if I want to find the acceptable values of ##V_o## for the bound states to be ##1,2,3,\dots## or none?